$\text{Ta có :} 3n-2 \vdots n+2$
$ 3n+6-8 \vdots n+2$
$ 3(n+2)-8 \vdots n+2$
$\text{mà } 3(n+2) \vdots n+2$
$⇒8\vdots n+2⇒n+2∈Ư_{(8)}=\text{{±1;±2;±4;±8}}$
lập bảng
$\left[\begin{array}{ccc}n+2&-8&-4&-2&-1&1&2&4&8\\n&-10&-6&-4&-3&-1&0&2&6\end{array}\right]$
$\text{Vậy n = {-10;-6;-4;-3;-1;0;2;6}}$